How to prove the concentration equality for standard normal?

Question

The following inequality is given in some of Yale's online lecture notes

$$P(|Z|>x) \leq 2 \sqrt{2 \pi} \phi(x)$$

Where $Z \sim N(0,1)$ with density $\phi(x)$. They call it a concentration inequality, the proof is not given.

I know how to prove the bounds

$$(1/x - 1/x^3)\phi(x) \leq 1-\Phi(x) \leq 1/x \phi(x)$$

But the method I used for these did not get me anywhere.

I know that it is enough to show

$$\Phi(x) \geq 1- \sqrt{2 \pi} \phi(x)$$

by symmetry of normal distribution. So we need to show

$$\int_x^\infty \phi(y)dy \geq 1- \sqrt{2 \pi} \phi(x)$$

Edit:

Following advice from answer

Note that $E[e^{Zt}] = e^{t^2/2}$ and we already have the inequality

$$P(Z>x) \leq E[e^{Zt}]/e^{xt}$$

Maximising the RHS equality with respect to $t$ gives $t_\max = x$, and gives the upperbound needed.

Answer 1

It is not true. When $x=0.1$, $P(|Z|>x) = 0.92$ and $2\phi(x)=0.79$.

Maybe the correct one is:

$$P(|Z|>x) \leq 2 \sqrt{2\pi}\phi(x)$$

For new inequality, following follow step:

1. Get $E(e^{\lambda Z})$

2. Find $\sup_\lambda(\lambda x - \log E(e^{\lambda Z}))$

3. Use Chernoff bound, you can get $\Pr(Z>x) < e^{-t^2/2}$